Examine if Rolle's Theorem is applicable to any of the following functions. Can you say some thing about the converse of Roller's Theorem from these examples?
$f(x)=[x]$ for $x \in[5,9]$
Examine if Rolle's Theorem is applicable to any of the following functions. Can you say some thing about the converse of Roller's Theorem from these examples?
$f(x)=[x]$ for $x \in[5,9]$
By Rolle's Theorem, for a function $f:[a, b] \rightarrow R,$ if
a) $f$ is continuous on $[a, b]$
b) $f$ is continuous on $(a, b)$
c) $f(a)=f(b)$
Then, there exists some $c \in(a, b)$ such that $f^{\prime}(c)=0$
Therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.
$f(x)=[x]$ for $x \in[5,9]$
It is evident that the given function $f(x)$ is not continuous at every integral point.
In particular, $f(x)$ is not continuous at $x=5$ and $x=9$
$\Rightarrow f(x)$ is not continuous in $[5,9]$
Also $f(5)=[5]=5$ and $f(9)=[9]=9$
$\therefore f(5) \neq f(9)$
The differentiability of $f$ in $(5,9)$ is checked as follows.
Let $\mathrm{n}$ be an integer such that $n \in(5,9)$
The left hand limit limit of $f$ at $x=n$ is.
$\mathop {\lim }\limits_{x \to 0'} \frac{{f(n + h) - f(n)}}{h} = \mathop {\lim }\limits_{x \to 0'} \frac{{[n + h] - [n]}}{h} = \mathop {\lim }\limits_{x \to 0'} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{x \to 0'} 0 = 0$
The right hand limit of $f$ at $\mathrm{x}=\mathrm{n}$ is,
$\mathop {\lim }\limits_{h \to {0^\prime }} \frac{{f(n + h) - f(n)}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{[n + h] - [n]}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{n - n}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} 0 = 0$
Since the left and right hand limits of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x=n$
$\therefore f$ is not differentiable in $(5,9).$
It is observed that $f$ does not satisfy all the conditions of the hypothesis of Rolle's Theorem.
Hence, Rolle's Theorem is not applicable for $f(x)=[x]$ for $x \in[5,9].$
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